# Find T with Subspace S as Kernel of T

1. Mar 24, 2012

### Bacle2

Hi, All:

I have been tutoring linear algebra, and my student does not seem to be able

to understand a solution I proposed ( of course, I may be wrong, and/or explaining

poorly). I'm hoping someone can suggest a better explanation and/or a different solution

to this problem:

We have two vector spaces V,V' , over the same F, and we have a subspace S of V.

The goal is to construct a map T , whose kernel is precisely S. The dimensions of

S,V,W respectively work well re Rank-Nullity, i.e., DimV-DimS=DimV'. My goal is to

declare T to be zero in B_S , and then set up a bijection between the remaining basis

vectors in B_V , and the basis vectors in B_V'.

So, I propossed that:

i)We choose a basis B_S :={e_1,...,e_s} for S, extend to a basis B_V:= {e_1,e_2,....,e_b}

for V. Let B_W:={e_1',e_2',...,e_w'} ; s:=|B_S|, and so-on.

ii)Declare/define T(B_S)==0 , i.e., for each basis vector e_s in B_S, we define

T(e_s)=0

iii) Now, we set up a bijection between the basis vectors in B_V\B_S , and those in

B_W. This bijection, gives rise to an isomorphism (extending by linearity) between

Span(B_V-B_S) , and V' , so we have:

1)T(e_s)==0 , for e_s in B_S

2)T(e_s+i):=e_i'

3)T(ce_s+i+de_s+j):=cT(e_s+i)+dT(e_s+j)

Now, I'm trying to extend this to the infinite-dimensional case, but my student has

only a beginners' knowledge of set theory.

Any Suggestions?

P.S: She also asked me a sort-of-strange question: is there such a thing as "non-linear

2. Mar 25, 2012

### chiro

Hey Bacle2.

For "non-linear algebra", there is an area of mathematics that is known as operator algebras and there are theorems that show how to given an operator O with certain constraints, you can find the definition for F(O). For example you can find the operator corresponding to SQRT(O) or e^(O) where O is a linear operator.

Also with regard to your 'extend to the infinite-dimensional case', things get really tricky because you can't just apply your finite-dimensional theorems to infinite-dimensional problems in general.

The reason has to do with the nature of infinity. Just like we need convergence for power series, we also need these kinds of properties for associates spaces, basis, and other things in infinite dimensional contexts. It's an absolute pain but it's got to do the fact that infinity just changes the game completely.

If you have a basis matrix for basis A and one for basis B then you derive the basis transformations using the fact that if z is the vector in the universal basis (orthonormal cartesian with a nxn identity matrix as basis) then we have

Uz = a
Vz = b where a is in basis A and b is in basis B with U being basis transformation for basis A and V is same for basis B. You can use the above to find information about going from one basis to another in terms of the actual matrix transformations and if we assume that U and V are proper bases then all the stuff you've talked about before (Rank nullity and so on) has to hold.

If you are talking about non-linear basis transformations you need to use the tensor identities for this, but I'm assuming that you are only refering to linear systems which means the above will hold.

In terms of the infinite-dimensional case, you will need to show specifically that properties of the norm will converge and this means that you will need to consider convergence for an 'infinite by infinite' kind of matrix. If you show that the appropriate convergence properties hold then you can show the appropriate identities for the infinite dimensional cases.

Also is something is a bijection, then the dimension of the basis should not change. If you are mapping say a vector x to y from space A to B and the space is linear in the way that we can represent things by a transformation matrix, then A and B will the same dimension and the basis have to also be the same dimension in terms of the operators involved.

3. Mar 25, 2012

### Bacle2

Thanks, Chiro; re non-linear algebra, now that you mention your examples, I think

the Borel Functional Calculus, where one designs a calculus for operators, where one

treats operators as numbers, i.e., we can define square roots, logs, etc. on operators.

Is that what you were referring to?

Re the infinite-dimensional case, you are right; I had not really considered the issue

of rank-nullity. Otherwise, the issue of bijection goes thru, but you're right, there are

4. Mar 25, 2012

### chiro

Yeah the functional analysis is where this kind of thing is considered, but there is a specific area of operator algebras which pays a lot of attention and studies these kinds of things specifically.

What happens is that topologies are brought in since they are one of the more general ways to study continuity and which end up being used to study convergence since if something doesn't converge, it doesn't hold any water no matter what structure, function, whatever we are talking about.

Also another thing is that for infinite-dimensional vectors, you have to impose restrictions on these in addition to the basis because we usually consider these things in a space (like little l^1 or l^2) and for something to be in this space, it has to satisfy a property with respect to some kind of norm in the way that the norm identity of the actual vector converges.

If you want to get a better idea of this, take a look at Hilbert spaces: this is kind of the motivation for all this infinite-dimensional stuff and one thing that is studied intensely are infinite-dimensional systems that appear in quantum mechanics: this is what Von Neumann was working on and its not surprising that he was actually working under Hilbert directly among others which no doubt, helped him build the first rigorous mathematical foundation for QM.

5. Mar 25, 2012

### chiro

Also before I forget, you should take into account the property of linearity especially for operators.

The main linearity conditions are basically f(aX+bY) = af(X) + bf(Y) and if you find any case where this is violated in terms of an operator, then you by definition have a non-linear operator.

6. Mar 25, 2012

### Bacle2

Thanks, Chiro, I figured out a way.